Chapter 2 Elementary Spectroscopy - CTAPSctaps.yu.edu.jo/...Phys645_Chapter2_Spectroscopy_C.pdf ·...

1

© Dr. Nidal M. Ershaidat

Phys. 645: Environmental Physics

Physics Department

Yarmouk University

Chapter 2 Chapter 2 Elementary SpectroscopyElementary Spectroscopy

http://ctaps.yu.edu.jo/physics/Courses/Phys645/Chapter2

© Dr. Nidal M. Ershaidat Environmental Physics - Chapter 2 Spectroscopy

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Outline

2-1 The Solar Spectrum2-2 Light-Matter Interaction2-3 Biomolecules, Ozone and UV LightAppendix: The Ozone Hole


3

Spectroscopy: Generalities 1

Measurements in environmental physics result generally in a spectrum.

The variation of the intensity (of light) with some variable (frequencies of this light) I = f(νννν) is called a spectrum.

Spectroscopy = Study of spectra.

Spectroscopist = physicist or scientist specialized in spectroscopy.


4

Spectroscopy: What is a spectrum?

Example 1Example 1: �When we represent the RADIANCY of a black body as a function of the frequencies emitted by this BB we obtain a “black body spectrum”

TT/2

Figure 2-1


5

Black Body - Definition

The Black Body radiation was one of the important problems in the physics of the 20th century.

A Black Body is an object which emits and absorbs perfectly all wavelengths of the electromagnetic spectrum.


6

Black Body Spectrum

The light emitted or absorbed of a black

body is very characteristic.

TT/2

2


7

Spectroscopy Example 2Example 2:

When we represent the INTENSITY of X-rays emitted by an element as a function of the frequencies we obtain the “X-ray spectrum”

Figure 2-2


8

2-1 The Sun Spectrum - Generalities

� Light reaching the earth from the sun

is essential for life.

� The detailed balance between the

inflow and the outflow of solar energy

establishes the temperature of the

earth’s surface.

� The solar light is absorbed by

photosynthetic organisms & energy

is conversed into chemical free energy

for later use


9

The Sun Spectrum vs. Earth

This is also true for Light re-emitted by the earth’s surface and atmosphere

Light arrives on the earth’s surface and it is composed of a wide range of frequencies characteristic of:

1) The emitter (the sun)2) Specific elements of the solar surface 3) The composition of the atmosphere responsible for the transmission of light to the earth’s surface


10

2-1 The Sun Spectrum & Ozone (O3)

Ozone absorbs all light with a wavelength less than 295 nm, the UV region.

� if the ozone layer decreases then, not only, UV will reach the earth’s surface but all shorter wavelengths.

And this is very dangerous for life.

�Biomolecules such as the DNA (carrier of genetic capital) and proteins are sensitive to photons

2-1 The Solar Spectrum


12

Figure 2-3: Absorption spectrum by various pigments

IR

The Sun & photosynthesis

UV

Pigment absorption

Wavelength (nm)

3

13

Pigment Absorption of the Sun light

Chlorophyll-a

• major pigment of higher plants, algae &

cyanobacteria• absorbs red and blue lightIn association with carotenoids they provide plants with their green color

Figure 2-3

Pigment absorption

Wavelength (nm)

Pigment is the�� the natural coloring matter of animal or

plant tissue.


14

Pigment Absorption of the Sun light

Pigment absorption

Wavelength (nm)

Bacteriochlorophyll-a & Bacteriochlorophyll-b are the major pigments absorbing in the near infrared region of the spectrum


15

Earth is an emitter (a black body: T=288K)

The earth does not only absorb and reflect light. It also emits light, and the emission can be considered as that of a Black Body at

T=288K. Its spectrum is in the far infrared.

The energy balance is based on the emission and partial absorption of this infrared light and any changes in any of these processes may disturb the balance.In particular, changes in CO2, known for its few strong absorption bands in the IR, may alter the mentioned above balance.


16

Environmental Physics &

Atomic and Molecular Spectroscopy

The goal of this chapter is to see how Atomic Spectroscopy and Molecular Spectroscopy are related to Environmental Physics


17

Spectroscopy gives answers

We need answers to several questions:

1- Why are specific colors of the light emitted

or absorbed by atoms and molecules?

2- Can we quantify the amount of absorption?

3- “Thickness” of an absorption band and its

implication?

4- Quantification of changes in the

atmospheric transmission spectrum due to

changes concentrations of CO2 and O3.


18

Answering the questions

Requires:

1- Basic understanding of the interaction of

light and matter (paragraph 2)

2- Knowledge of the terminology of

spectroscopy

4

2-2 Light-Matter Interaction


20

Introduction

In all these processes energy may be absorbed or released.

Interaction of Light with Matter = Interaction of an electromagnetic wave with matter.

When light enters a medium, it interacts

with the atoms and molecules of the medium and several processes may

occur.


21

The Frame: Quantum Mechanics

� Einstein Coefficients,� Optical Density & Lambert-Beer’s Law,�Application to the Ozone Layer.

Quantum Mechanics is the frame Quantum Mechanics is the frame

necessary to study such an necessary to study such an

interaction.interaction.


22

The Perturbation Theory

2-2-1 The Transition (Electric) Dipole Moment

Consider a quantum system, atomic or Consider a quantum system, atomic or

molecular system, with discrete energy levels molecular system, with discrete energy levels

EEkk and wave functions and wave functions 0

kψψψψ

are the unperturbed solutions of the timeare the unperturbed solutions of the time--

independent Schrindependent Schröödinger equation dinger equation

0

kψψψψ

00

0 kkkEH ψψψψψψψψ ==== 2-1


23

The Hamiltonian H0

For simplicity we shall consider that the

eigenvalues of H0 (i.e. Ek) are not

degenerated

The eigenstates of H0, are stationary. (((( ))))0

kψψψψ


24

Applying a time-dependent perturbation

At this instant At this instant tt, The state of the system will be, The state of the system will be

ΨΨΨΨ t

H1(t)will mix with H0, and the total hamiltonian

at an instant t is:

H(t) = H0 + H1(t)

Consider now that a timeConsider now that a time--dependent dependent

perturbation is applied on the system and that perturbation is applied on the system and that

the perturbation hamiltonian is the perturbation hamiltonian is HH11((tt))..

When the perturbation is applied, starting from

t=0, the system evolves.

Generally, the eigenstates are no more

eigenstates of the hamiltonian H(t).

(((( ))))0

kψψψψ

5


25

Transition Probability

Calculation of the probability Pkn that the

system be in some stationary state (((( ))))0

nψψψψ

is called the matrix element

ΨΨΨΨψψψψ tn

0

Transition Probability & Schröödinger

To calculate Pkn we need and this

means that we need to solve the Time-

dependent Schröödinger equation

(TDSE)

ΨΨΨΨ t


26

(((( ))))(((( ))))

ΨΨΨΨ++++====∂∂∂∂

ΨΨΨΨ∂∂∂∂ttHH

t

ti 10�

Time-dependent SchrSchröödinger equation dinger equation

verifies the TDSEverifies the TDSE

verifies the initial condition: verifies the initial condition:

ΨΨΨΨ t

00

kt ψψψψ========ΨΨΨΨ

is a unique solution of is a unique solution of Eq. 2Eq. 2--22..

ΨΨΨΨ t

ΨΨΨΨ t

2-2


27

in the eigenfunctions basein the eigenfunctions base

can be written as a linear combination of can be written as a linear combination of

the base of eigenfunction the base of eigenfunction

ΨΨΨΨ t0kψψψψ

(((( )))) �/tEi

k

k

kketct

−−−−ψψψψ====ΨΨΨΨ ∑∑∑∑

0

ΨΨΨΨ t

(((( ))))

ΨΨΨΨΨΨΨΨ==== ttckk

0

2-3

2-4


28

�� h s of TDSE

(((( ))))

(((( ))))

(((( ))))∑∑∑∑

∑∑∑∑

∑∑∑∑

−−−−

−−−−

−−−−

ψψψψ++++

ψψψψ

∂∂∂∂

∂∂∂∂====

ψψψψ∂∂∂∂

∂∂∂∂

k

/tEi

k

k

/tEi

k

/tEi

k

kk

k

kk

k

kk

etcE

etct

i

etct

i

�

�

�

�

�

0

0

0

2-5


29

�� h s of TDSE by

(((( )))) �/tEi

kk

kkn

ketcE

−−−−∑∑∑∑ ψψψψψψψψ++++00

(((( ))))

(((( ))))

ψψψψψψψψ

∂∂∂∂

∂∂∂∂====

ΨΨΨΨ∂∂∂∂

∂∂∂∂ψψψψ

∑∑∑∑−−−−

k

/tEi k

knk

n

etct

i

tt

i

��

�

00

0

0

nψψψψ

2-6


30

2-7

r h s of TDSE

(((( ))))(((( )))) (((( ))))

(((( ))))

(((( )))) (((( ))))[[[[ ]]]]

ψψψψ++++

ψψψψ

====

ψψψψ++++

∑∑∑∑

∑∑∑∑

∑∑∑∑

−−−−

−−−−

−−−−

n

/tEi

k

/tEi

k

k

/tEi

n

nn

k

kk

k

kk

etHtc

etcE

etctHH

�

�

�

0

0

0

1

10

6


31

r h s of TDSE by

(((( ))))(((( )))) (((( ))))

(((( ))))

(((( )))) (((( ))))(((( )))) (((( ))))∑∑∑∑

∑∑∑∑

−−−−

−−−−

ΨΨΨΨψψψψ++++

ψψψψψψψψ

====ΨΨΨΨ++++ψψψψ

n

/tEi

k

/tEi

k

n

nn

k

knk

n

ettHtc

etcE

ttHH

�

�

1

0

00

10

0

n

0ψψψψ

(((( )))) (((( ))))tcWetct

i nnkn

t

k

nk∑∑∑∑

ωωωω

====∂∂∂∂

∂∂∂∂ i

�

Multiplying by Multiplying by �/tEi ne

−−−−

2-8

2-9


32

Finding ck(t)

wherewhere

(((( )))) (((( ))))tcWei

tct

nnkn

t

k

nk∑∑∑∑

ωωωω−−−−====

∂∂∂∂

∂∂∂∂ i

�

�

nknk

EE −−−−====ωωωω0

1

0

nknk HW ψψψψψψψψ====

2-10

2-11andand

We can also use the notation:We can also use the notation:

nHkW nk 1==== 2-12


33

Using initial conditions

Taking Taking

Then at Then at tt = 0= 0

cc11(0) = 1(0) = 1 and and cckk(0) = 0(0) = 0 (for (for kk >1>1) )

Then for a time Then for a time tt sufficiently short sufficiently short

cc11((tt) ) ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ 11 and and cckk((tt) ) ≈≈≈≈≈≈≈≈ 00 (for (for kk >1>1) )

(((( )))) 0

10 ψψψψ========ΨΨΨΨ t 2-13

34

Computing ck(t)

Integrating Eq. 2-10 in these conditions*:

yields:

(((( )))) (((( ))))tcWei

dt

tdcnnk

n

tk nk

∑∑∑∑ωωωω

−−−−====i

�

(((( )))) tdWei

tc k

t t

k

k ′′′′−−−−==== ∫∫∫∫′′′′ωωωω

1

0

1i

�

where 11

0

11

0

1HkHW kk ====ψψψψψψψψ====

(((( ))))�

1

1

EEkk

−−−−====ωωωωand

2-14

2-15

2-16b

*Note that Eq. 2-14 is an ODE and no more a PDE.

2-16a


35

Probability of “the excited state k”

The probability of finding the quantum system

in an excited state k, thanks to the absorption

of a quantum of energy is (((( )))) (((( ))))2

tctkk ====P

This probability is related directly

2

1

2

11HkWk ==== 2-17

36

An incident light, (a light source is switched on) i.e.

an incident electromagnetic wave, with frequency

ωωωω will “perturb” the system, and a perturbation

hamiltonian H1 will mix with the hamiltonian of

the system H0.

Interaction of Light with Matter=Perturbation

The atomic or molecular system is considered at the origin of the coordinates system, and being globally neutral it will interact with the electric

component of the em wave through a moment µµµµ.

for simplicity we shall consider a linearly polarized light and neglect the magnetic component of the em wave (Why can we do that?).

(((( )))) tcosEtE ωωωω==== 0

7


37

Perturbation = A linearly polarized light

The perturbation hamiltonian is:

(((( )))) (((( ))))tEtH��

••••µµµµ−−−−====1 2-18

∑∑∑∑====µµµµi

i ii rq��

where is the electric dipole operator: µµµµ�

The sum is taken over all electronic charges qi at position ir

�

(((( )))) θθθθωωωωµµµµ==== costcosEtH01 ⇒⇒⇒⇒ 2-19

(((( ))))0E,��

µµµµ====θθθθ^


38

PPkk

(((( )))) tdWei

tct

k

ti

k

k ′′′′−−−−==== ∫∫∫∫′′′′ωωωω

0

1

1

�

θθθθωωωωµµµµ====→→→→

costcosEkWk 011

∫∫∫∫ ′′′′′′′′ωωωωθθθθµµµµ−−−−

====′′′′ωωωω−−−−

→→→→ tti

tdtcosecosEki

k

0

1

01

�

(((( )))) (((( ))))

(((( ))))2

1

12

1222

02

21

1

2

ωωωω−−−−ωωωω

θθθθµµµµ====

====

ωωωω−−−−ωωωω

k

tk

kk

sincosEk

tctP

�

�

⇒⇒⇒⇒

2-20

2-21

2-22


39

ComputingComputing

(((( ))))∫∫∫∫ ′′′′′′′′ωωωω

′′′′ωωωω−−−−′′′′ωωωω====

tkk

tdtcostsinitcos

0

11

2

∫∫∫∫ ′′′′′′′′ωωωω′′′′ωωωω−−−−

tt

ki

tdtcose0

1

∫∫∫∫ ′′′′′′′′ωωωω′′′′ωωωω−−−−

tt

ki

tdtcose0

1


40

The probability that the system (atomic or

molecular) which was initially (at t = 0) in the

state 1 be in the state k at t is thus proportional to the (square of the matrix element )

Transition dipole momentTransition dipole moment

(((( ))))22

11

µµµµ====µµµµ∝∝∝∝ ktkk

P

(((( ))))2

11

2µµµµ====µµµµ kt

k transition dipole momenttransition dipole moment

For an isotropic solution or a gas, we should average all over the angles θ θ θ θ

2-23


41

Transition dipole moment rateTransition dipole moment rate

The transition probability peaks sharply around ωωωω = ωωωωk1

For a non monochromatic light, one should average over a “frequency band” and this leads to the rate of population of level k:

(((( )))) (((( )))) (((( ))))ωωωω≡≡≡≡ωωωωµµµµεεεε

ππππ==== WBW

dt

tdkk

k1

2

1

03 �

P2-24

W(ωωωω) = Time averaged energy density of the incident em field at frequency ωωωω.

The Einstein CoefficientsLambert-Beer’s Law

8


43

Einstein CoefficientsEinstein Coefficients

In this expression we have averaged over all possible angles between E0 and µµµµ

2

1

0

1

3kk

B µµµµεεεε

ππππ====

�

Bkk11 is the Einstein Coefficient for absorption and stimulated emission is directly related to the extinction coefficient measured in an absorption experiment.


ππππ==== WBW

dt

tdkk

k11

2

03 �

P

W(ωωωω) = Time averaged energy density of the incident em field at frequency ωωωω.

2-25


44

Einstein CoefficientsEinstein Coefficients

Here, we shall establish a simple relation between the rates of absorption, stimulated emission and spontaneous emission in an atomic or molecular system

Consider a quantum system, and two energy levels E1 & E2. Consider that the population of

these 2 levels are respectively N1 & N2.

Radiative Processes


46

Transitions between 2 energy levelsTransitions between 2 energy levels

3 possible transitions (called radiative

processes) are possible between E1 & E2.

Absorption: Level 1 absorbs incident light (frequency ωωωω) and the system

“transits” to level 2.

B12 W(ωωωω)AbsorptionLight is ON

E2, N2

E1, N1

Figure 2-4-a


47

Stimulated emissionStimulated emission

Stimulated emission: Level 2 emits light (frequency ωωωω) and the system “transits”

to level 1.

StimulatedemissionLight is ON

E2, N2

E1, N1

B21 W(ωωωω)

Figure 2-4-b


48

Spontaneous emissionSpontaneous emission

Spontaneous emission: Level 2 emits light (frequency ωωωω) spontaneously and

the system “transits” to level 1.

SpontaneousemissionLight is OFF“Dark Process”

E2, N2

E1, N1

A21 W(ωωωω)

Figure 2-4-c

9


49

BB1212 = = BB2121 & & AA2121 = = aa BB2121

The rate of population of level 1 is given by:

For a steady state (((( ))))01 ====

dt

tdN

(((( ))))(((( ))))

2112

21

21 BNNB

AW

−−−−====ωωωω

(((( ))))221

1

212112NANWBNWB

dt

tdN++++++++−−−−====

Light is off

2-26

2-27


50

Ratio Ratio NN11//NN22

Tke

N

N ωωωω==== �

2

1

In the absence of an external radiation field

and for a system in thermal equilibrium the

ratio N1/N2 follows Boltzmann Distribution:

eVeV.Tk40

10250 ====≈≈≈≈

k = Boltzmann constant = 1. 38 x 10-23 J K-1

and for T = 300 K (Room Temperature):

2-28

2-29


51

The energy density The energy density WW((ωωωωωωωω))

The energy density W(ωωωω) is given by Planck’s radiation law

(((( ))))1

132

3

−−−−ππππ

ωωωω====ωωωω

ωωωω TkecW

�

�

The 2 expressions are identical and this gives:

(((( ))))(((( )))) 1

2112

2121

−−−−====ωωωω

ωωωω TkeBB

BAW

�

2-30

2-31

We have seen (eq. 2-27) that:


52

Relation between Einstein CoefficientsRelation between Einstein Coefficients

B12 = B21

12

21

32

3

21

21

B

A

cB

A====

ππππ

ωωωω====�

2

21

0

32

21

0

3

3

2133

µµµµππππ

µµµµωωωω====µµµµ

εεεεππππ

ωωωω====

ccA

2

21

0

21

3µµµµ

εεεε

ππππ====

�B

12

00 ====µµµµεεεε cusing

2-32

2-33


53

||µµµµµµµµ1212||22 determines the rates of determines the rates of

all radiative processesall radiative processes

Major conclusion:

||µµµµµµµµ1212||22 determines the rates of all radiative determines the rates of all radiative

processes and consequently common processes and consequently common

selection rules apply to these processesselection rules apply to these processes


54

LambertLambert--BeerBeer’’s Laws Law

How are the macroscopic quantities (we measure) related to the “quantum”

Einstein coefficients and µµµµ12?

10


55

Macroscopic absorptionMacroscopic absorption

A classical absorption measurementA light of intensity (flux) I with “band width” dωωωω falls on a sample of thickness dz.

Light is absorbed especially when the

light is weak and the atoms / molecules

of the absorber are in their ground state

N1 >> N2 (except in the case of laser where

it is amplified)


56

The rate of transitions between states 1 & 2

This rate is given by: (No stimulated emission)

(((( )))) (((( )))) 21

1

2112NANWB

dt

tdN++++ωωωω−−−−====

B12N1 W(ωωωω) = A21 N2

For a steady state(((( ))))

01 ====dt

tdNLight is off

2-34

2-35

Eq. 2-35 is the rate at which energy is removed

from the incident beam or the Intensity loss rateIntensity loss rate.


57

Classical Macroscopic absorptionClassical Macroscopic absorption

Light passes through a thickness dz. (The total volume of the absorber is V)

dωωωωI dωωωω

dz

∂∂∂∂

∂∂∂∂++++

z

II

W dωωωω Area a

Figure 2-5


58

““Energy lossEnergy loss””

Because of attenuation W is a function of z

The amount of energy in the frequency interval [ωωωω, ωωωω+dωωωω] through the slice (a dz) is W dωωωω a dz. (J. s-1)

The rate at which the beam’s energy decreases is:

(((( ))))dzad

t

tWωωωω

∂∂∂∂

∂∂∂∂−−−−

131

3

−−−−−−−−−−−−

==== sWattmss

mJ

2-36


59

All atoms do not absorb at the same frequencyAll atoms do not absorb at the same frequency

First we should take into account the fact that the atoms/molecules of the system do not absorb light at exactly the frequency ωωωωk.

(((( )))) 1====ωωωωωωωω∫∫∫∫ dF

We define the fraction of transitions that occur in the interval [ωωωω+dωωωω]: F(ωωωω)

dωωωω with the normalization condition:

2-37


60

(((( ))))dzad

t

tWωωωω

∂∂∂∂

∂∂∂∂−−−−====

(((( )))) (((( )))) ωωωωωωωωωωωωωωωω dFV

dzaWBN �121

Energy balanceEnergy balance

The rate of decrease of the beam energy = the rate at which the light energy is removed from the beam by absorption. This implies:

From (Eq. 2-35)

11


61


(((( )))) (((( ))))ωωωωωωωω

ωωωω−−−−====∂∂∂∂

∂∂∂∂⇒⇒⇒⇒ F

VWBN

t

W �121

In other words:

z

I

t

W

∂∂∂∂

∂∂∂∂−−−−====

∂∂∂∂

∂∂∂∂−−−−

Energy entering the slice (adz) - Energy leaving it per time unit = rate of decrease of beam energy.

2-39

2-38

Where is the change in intensity (in W.m-2) of

the beam upon passage through the slice a dz

z

I

∂∂∂∂

∂∂∂∂−−−−


62

W(W(ωωωωωωωω) and the optical properties of the absorber) and the optical properties of the absorber

The energy density distribution W(ωωωω) according to optics equals:

nc

IW ====

n is the index of refraction for the absorber

2-40


63


I(z) = I0 e- K z Lambert-Beer’s Law

Integrating eq. 2-41 gives:

(((( ))))IKI

ncV

FBN

z

I====

ωωωωωωωω−−−−====

∂∂∂∂

∂∂∂∂ �121

K is the (linear) absorption coefficient [K] = L-1

(((( ))))ncV

FBNK

ωωωωωωωω====

�121

2-41

2-42

2-43


64


ππππ==== WBW

dt

tdkk

k11

2

03 �

P

B21 is given by (F(ωωωω) is normalized) :

∫∫∫∫ωωωω

ωωωω−−−−====

band

dK

N

ncVB

�12

Computing Einstein Coefficients

Integrating K, which is the quantity we measure in an absorption experiment, over the

absorption line gives B21 and consequently B12

and A21 and |µµµµ12|2

2-44


65

I = I(0) 10-OD

OD = Optical Density = εεεε C l

εεεε = molar extinction coefficient (in dm3 mole-1 cm-1 )

C = concentration of the absorber sample(in mole dm-3)

l = pathlength (in cm)

Optical DensityIn practice, the dependence of I on passage through a

material of length l is expressed as follows:

2-45


66

Optical Density for chlorophyll a

C = concentration = 10-3 mole-1 dm3

l = pathlength = 0.02 cm

At its absorption maximum, λλλλ = 680 nm

(red light), chlorophyll a has:

εεεε = 105 dm3 mole-1 cm-1

OD = 2

12


67

Chlorophyll a absorbs red light

OD = 2 gives a reduction of the intensity of light

of the order of 100!!

All red light is absorbed by a leaf of chlorophyll a


68

The transition dipole moment

2

1

0

1

3kk

B µµµµεεεε

ππππ====

�

From the relation between the transition dipole moment and Einstein coefficient B21

The most important result


ωωωω

εεεε

ππππ====µµµµ

band

dK

N

ncV

��0

2

123


ωωωωεεεε××××====µµµµ −−−−

band

d.

612

12 10011

Check this!

2-46


69

For a single molecule, the absorption cross section is given by:

σ σ σ σ = ππππ r2 P

r = radius of the moleculeP = Probability that the photon is

absorbed by the surface ππππ r2

The extinction coefficient ε ε ε ε

30323032

2

.

Nr

.

N AA ππππ====

σσσσ====εεεε See Chapter 4

NA is Avogadro’s number.

2-3 Biomolecules, Ozone & UV Light


71

• Absorption of UV light in biological molecules

• Importance of the ozone layer

• Effects of the deterioration of the Ozone layer and Lambert-Beer’s Law

Section 2-372

• Biomolecules

The most sensitive to UV are:

The Spectroscopy of Biomolecules

•Nucleic Acids DNA & RNA

DNA (Deoxyribonucleic acid) is the carrier of the genetic code. It is responsible of Reading and translating the genetic code

• Essentially proteins: such as the αααα-

crystallin which is the major protein of the eyelens.

13

73

The UV region

• UV-A (near UV region)

320 nm < λλλλ < 400 nm

• UV_B (mid UV region)

290 nm < λλλλ < 320 nm

• UV-C (far UV region)

200 nm < λλλλ < 290 nm

Absorption of Light by Biomolecules


75

Absorbance - DNA and αααα-crystallin

Figure 2-6: Absorption spectra of DNA (- • • • • - • • • •) and αααα-crystallin (--) in

the wavelength region 240-340 nm.

DNA

Solar spectrum


76

Absorption of Light by DNA

For DNA, The spectrum peaks at λλλλ = 260 nm with

a maximum extinction εεεε = 104 dm3 mole-1 cm-1

Fig. 2-6 shows the UV solar spectrum and the absorption by DNA and αααα-crystallin


77

Aromatic DNA bases

Absorption of light by DNA is due to the aromatic DNA bases: guanine, thymine, cytosine and adenineThe electronic transitions which contribute to the absorption in the range 220-290 nm are predominantly oriented in the plane of the DNA base!

For a typical cell, the absorption of solar UV due to protein would be about 10% of the absorption of nucleic acids.

See paragraph 2, page 24, for eventual absorption of light in the near UV region by proteins

Comparison


80

Comparing to the DNA absorption spectrum, the action spectrum is greater in the region

λλλλ=320 nm. Other absorbers exist but the most sensitive is the DNA.

How damage is done?

It is the chromophores of DNA that form the prime target in the process that leads to the photodamage.

Interaction of light with thymine, or a cytosine-thymine pair produces the pyrimidine (photo product) which alters the reading and translation of the genetic code!

14


81

Quantifying the Photodamage

The damage done to a biological system by solar UV can be calculated from the action spectrum!

E(λλλλ) has a large negative slope, I(λλλλ) has a

positive slope.

D is sensitive to even slight changes in I(λλλλ).

(((( )))) (((( ))))∫∫∫∫∞∞∞∞

λλλλλλλλλλλλ====0

dIED 2-47

22--33--3 The Ozone filter3 The Ozone filter


83

The Ozone Filter

aa-- What is the Ozone

layer?What is th

e Ozone layer?

bb-- How is Ozone form

ed?How is Ozo

ne formed?

cc-- How does Ozone f

ilter the solar

How does Ozone filte

r the solar

spectrum?spectrum?

dd-- What damages the

ozone layer?

What damages the o

zone layer?

The Ozone holeThe Ozone hole


84

� O3 forms a thin layer in the stratosphere.

� Most atmospheric Ozone is in the

stratosphere, about 15-30 km.

� The maximum concentration is found

between 20 & 26 km above the earth’s

surface.

a-What is the Ozone layer

�A concentration of ozone molecules in the atmosphere of the earth. The amount of O3corresponds to a layer of 0.3 cm under NTP.


85

a-What is the Ozone layer

�Ozone

Figure 2-7© Dr. Nidal M. Ershaidat Environmental Physics - Chapter 2 Spectroscopy

86

The Earth’s Atmosphere

�The Troposphere:

- The lowest region. It extends from the

earth’s surface up to about 10 km in altitude.

- All human activities occur in the troposphere.

It is divided in several layers

�The stratosphere:

- continues from 10 km to 50 km.

- Commercial airline traffic occurs in the lower part of the stratosphere.

15


87

Mesosphere, Thermosphere & Exosphere

�The mesosphere:


�The Thermosphere:


�The Exosphere:



88

Ozone in the Atmosphere

Figure 2-8


89

b- (How is) The Ozone O3 (formed?)

Ozone is a molecule containing 3 oxygen atoms

while normal oxygen (which we breath) is colorless & odorless, Ozone is blue in color and has a strong odor

� O3 is formed through the combination of O2 molecule with an Oxygen atom:

�O2 + O O3


90

Ozone is rare in our “air”

Out of 107 molecules of air: 2 million are normal Oxygen for 3 Ozone molecules!!

The ozone layer absorbs the UV-B portion of the sun spectrum.

The Ozone O3 (Properties)

The UV-B is known to be the major cause of skin cancer and cataracts. It is also harmful to some materials and to the marine life!


91

Formation and destruction of ozone molecules take place all the time, but its concentration remains relatively stable.

Until recently, the variation (reduction and increase) of the ozone layer, due to sunspots, seasons and according to the altitude, remains nil.

Concentration of Ozone changes!


92

c- Formation of the atmospheric Ozone

�Where do O2 molecules and Oxygen

atoms, in the stratosphere, come from?

� The atomic oxygen is formed by photodissociation (dissociation by light) of the oxygen molecules in the 100 km region by wavelengths < 170 nm.

16


93

Photodissociation of O2(λλλλ < 175 nm)

�Step 1

� The oxygen molecule in its ground state ΣΣΣΣu-

interacts with the λλλλ < 175 nm photon and transits to the excited state ΣΣΣΣg

- )

�λλλλ < 175 nm

∑∑∑∑−−−−

u

3

∑∑∑∑−−−−

g

3

Figure 2-9a

© Dr. Nidal M. Ershaidat Environmental Physics

94


�O (1D)

�O (3P)�2 eV

� O2 (excited state)

� The excited molecule dissociates to two oxygen atoms, one in the ground state (3P) and the other one in the excited state (1D)

Figure 2-9b

�Step 2


95


� A much weaker absorption (σσσσ=10-23 cm2 = 10

barn) of light in the λλλλ< 240 nm region occurs and

the resultant oxygen atoms are in the ground state (1P)

Units of cross-section at the nuclear level[1 barn = 10-24 cm2)]

For comparison σσσσRutherford = 1 barn


96

What does the Formation of ozone depend on?

The efficiency of formation of ozone by UV

depends on:

1. Availability of oxygen in the stratosphere

2. Changes in the Temperature of the stratosphere and its chemical “contents”

3. Dust from volcanic eruptions


97

Characteristics of the Ozone layer

�Because the equator is the region the most exposed to sunlight, ozone above the equator reaches a maximum.

�Ozone diffuses towards the poles where the layer it forms is larger than above the equator.

�Its concentration is highest in late winter and early spring


98

c- How does the ozone filter the solar spectrum?

The atmospheric Ozone absorbsessentially all wavelengths below 295 nm

almost 90-95% of solar UV.

17


99

The Ozone (absorption) spectrum

Figure 2-10

AppendixThe Ozone Hole

See http://www.theozonehole.com/


101

There are two pathways for the destruction of ozone

O + O3 2 O2O3 + O3 3 O2

Ozone depletion reaction

d- The Ozone hole

What destroys the Ozone layer?

These reactions are the net result of more complex reactions catalyzed by various gases and radicals


102

Radicals: Cl, NO and OH: The ODS

Chlorine: produced in the volcanic eruptions & in the CFC’s and HCFC’s

ODS = Ozone-Depleting Substance

CFC = ChlorofluorocarbonsHCFC’s = Hydrochlorofluorocarbonsused in refrigerators, as “propellants”and foam-blowing agents (sprays)


103

ODS: Hydroxyl radical OH & Nitric oxide

OH results from the breakdown of H2O

vapor

Source: HSource: H22O from supersonic airplanes!O from supersonic airplanes!

NO is formed by photodecomposition of

N2O - used as a fertilizer!-

Cl, NO & OH are responsible for the

destruction of 99% of the stratospheric

ozone!!


104

The CFC’s (An ODS)

Figure 2-11

18


105

The Ozone tragic cycle

Figure 2-12


106

Processes that determine ozone concentration

Figure 2-13


107

The aerosol UV absorbent spectrum

TOMS: Total Ozone Mapping Spectrometer

Figure 2-14


108

Is the Ozone depletion dramatic?

10% of ozone depletion will result in 45% increase of solar UV-B reaching the earth’s surface.

That’s why it is so important to monitor the ozone layer and study the “bad”effects of UV on life.


109

The Montreal Protocol

� Production of CFC’s is banned since

1995 and fully stopped since 2000

�As for HCFC’s: Production will be banned in 2020.

�(Methyl bromides: 2005)


110

Montreal Protocol - Summary of Control Measures

Total phase out by 2005

Total phase out by 201570% reduction by 2000

20% reduction by 200550% reduction by 2001

1995-1998 base level25% reduction by 1999

Freeze in 2002 at averageFreeze in 1995 at 1991 base levelMethyl bromide

Phased out end of 1995Phased out end of 1995Hydrobromofluorocarbons (HBFCs)

Total phase out by 2040Total phase out by 2020

at 2015 base level90% reduction by 2015

Freeze in 201665% reduction by 2010

35% reduction by 2004

Freeze from beginning of 1996Hydrochlorofluorocarbons (HCFCs)

Total phase out by 2015Phased out end of 1995Methyl chloroform

Total phase out by 2010Phased out end of 1995Carbon tetrachloride

Total phase out by 2010Phased out end of 1993Halons

Total phase out by 2010Phased out end of 1995Chlorofluorocarbons (CFCs)

Developing CountriesDeveloped CountriesOzone Depleting Substances

19


111

The Kyoto Protocol

The Kyoto Protocol is an international agreement

linked to the United Nations Framework Convention on

Climate Change.

The major feature of the Kyoto Protocol is that it sets

binding targets for 37 industrialized countries and the

European community for reducing greenhouse gas

(GHG) emissions.

These amount to an average of five per cent against

1990 levels over the five-year period 2008-2012. The

Protocol commits industrialized countries to stabilize

GHG emissions.


112

The Kyoto Protocol

The Kyoto Protocol was adopted in Kyoto, Japan, on 11 December 1997 and entered into force on 16 February 2005. The detailed rules for the implementation of the Protocol were adopted at COP 7 in Marrakesh in 2001, and are called the “Marrakesh Accords.”

Recognizing that developed countries are principally responsible for the current high levels of GHG emissions in the atmosphere as a result of more than 150 years of industrial activity, the Protocol places a heavier burden on developed nations under the principle of “common but differentiated responsibilities.”

See: http://www.alternate-energy-sources.com/Kyoto-Protocol-summary.html

© Dr. Nidal M. Ershaidat

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